This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: SOLUTIONS TO ASSIGNMENT 5 – ACTSC 431/831, FALL 2008 1. (a) We have E ( S ) = E ( N ) E ( X ) = 10 and Pr { S = 1 } = Pr { S = 3 } = 0 , Pr { S = 0 } = Pr { N = 0 } = 0 . 1 , Pr { S = 2 } = Pr { N = 1 ,X 1 = 2 } = 0 . 04 , Pr { S = 4 } = Pr { N = 1 ,X 1 = 4 } + Pr { N = 2 ,X 1 = 2 ,X 2 = 2 } = 0 . 128 . Thus, we have E [( S 5) + ] = E ( S ) E ( S ∧ 5) = 10 4 X x =1 x Pr { S = x }  5 1 4 X x =0 Pr { S = x } ! = 5 . 748 . (b) We have E [( S 4 . 6) + ] = E ( S ) E ( S ∧ 4 . 6) = 10 4 X x =1 x Pr { S = x }  4 . 6 1 4 X x =0 Pr { S = x } ! = 6 . 041 . (c) We have E [( S 4) + ] = E ( S ) E ( S ∧ 4) = 10 3 X x =1 x Pr { S = x }  4 1 3 X x =0 Pr { S = x } ! = 6 . 48 , V ar ( S ) = E ( N ) V ar ( X ) + V ar ( N )( E ( X )) 2 = 40 , and E [( S 4) 2 + ] = ∞ X x =5 ( x 4) 2 Pr { S = x } = E [( S 4) 2 ] 4 X x =0 ( x 4) 2 Pr { S = x } = V ar ( S ) + ( E ( S )) 2 8 E ( S ) + 16 4 X x =0 ( x 4) 2 Pr { S = x } = 74 . 24 . Hence, V ar [( S 4) + ] = E [( S 4) 2 + ] ( E [( S 4) + ]) 2 = 32 . 25. 2. (a) We have E ( N ) = 15 and E [( X i 25) + ] = E ( X i ) E ( X i ∧ 25) = 20 Z 25 e x/ 20 dx = 20 e 5 / 4 = 5 . 7301 . Thus, the net reinsurance premium for the excessofloss reinsurance is E ( N ) E [( X i 25) + ] = 85 . 95 . 1 (b) The survival function of S is 1 F S ( x ) = 15 16 e x 320 , x ≥ . Thus, the net reinsur ance premium for the stoploss reinsurance is E [( S 150) + ] = E ( S ) E ( S ∧ 150) = E ( N ) E ( X i ) Z 150 15 16 e x/ 320 dx = 300 e 15 / 32 = 187 . 735 ....
View
Full
Document
This note was uploaded on 02/02/2010 for the course ACTSC 331 taught by Professor David during the Fall '09 term at Waterloo.
 Fall '09
 david

Click to edit the document details